Question

consider a sample of 53 football games, where 28 of them were won by the home team. use a 0.10 significance level to test the claim that the probability that the home team wins is greater than one - half. identify the null and alternative hypotheses for this test. choose the correct answer below. a. ( h_0:p = 0.5 ) ( h_1:p>0.5 ) b. ( h_0:p>0.5 ) ( h_1:p = 0.5 ) c. ( h_0:p = 0.5 ) ( h_1:p<0.5 ) d. ( h_0:p = 0.5 ) ( h_1:p
eq0.5 ) identify the test statistic for this hypothesis test. the test statistic for this hypothesis test is 41 (round to two decimal places as needed.) identify the p - value for this hypothesis test. the p - value for this hypothesis test is (round to three decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 28$ (number of home - team wins) and $n=53$ (sample size). So, $\hat{p}=\frac{28}{53}\approx0.5283$.

Step2: Calculate the test statistic

The formula for the test statistic $z$ in a one - sample proportion test is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$, where $p = 0.5$ (under the null hypothesis).
Substitute the values: $z=\frac{0.5283 - 0.5}{\sqrt{\frac{0.5\times(1 - 0.5)}{53}}}=\frac{0.0283}{\sqrt{\frac{0.25}{53}}}=\frac{0.0283}{\sqrt{0.004717}}\approx0.41$ (already given).

Step3: Calculate the P - value

Since this is a right - tailed test ($H_1:p>0.5$), the P - value is $P(Z>z)$.
We know that $z = 0.41$. Using the standard normal distribution table or a calculator, $P(Z>0.41)=1 - P(Z\leq0.41)$.
From the standard normal table, $P(Z\leq0.41)=0.6591$.
So, $P(Z>0.41)=1 - 0.6591 = 0.341$.

Answer:

The P - value for this hypothesis test is $0.341$.